# Duration

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```					                                  Duration
 While you receive the whole present value (price) of a zero at the date of
its maturity, with a coupon bond you receive part of the present value
before maturity.
 The duration of a bond measures the average lenght of time you have to
wait until receiving its whole present value.
 Assume a flat term structure. The price of a bond paying x1 , x 2 , , x n at
t  1,2 n is:
x1     x2           xn
[l] P                     

1  R 1  R 2        
1  R n
 Macaulay (NBER, 1938) defined the bond’s duration as:
x 1  R                 x 1  R                 x 1  R
1                      2                         n
[2] D  1  1                     2 2                  n  n
P                         P                         P
i.e. as the weighted average of the waiting times until cash payments are
made. The weights are given by the fractions of the total present value to
be received at t  1,2n . For a zero: D=n; for a coupon bond D<n; for a
perpetuity it can be shown that: D  1  R / R .
 The duration of a bond is important because it represents a measure of
the sensitivity of the bond’s price to interest rate changes. By taking the
derivative of [1] with respect to R, we obtain:
P
  x1 1  R  2 x 2 1  R  3x 3 1  R  x n 1  R
2              3             4                n 1
R

P
R

1
1 R

1x1 1  R  2x 2 1  R  3x 3 1  R  x n 1  R
1             2             3               n

and, with [2], we get:
P     1
[3]             D P     or                                                approximately:
R    1 R
P     D
      R   DR
P        1 R
which shows that an increase of 1 basis point in R (say from 3.00% to
3.01%) will cause a drop in the bond price of D basis points (0.10% if
D=10).

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 For values of R which are not very small, instead of D, it is better to use
the modified duration: D/1+R.
D
 It is possible to show that:  0 and, for bonds with fixed coupons c,
R
D                 D
that:     0 . Moreover:      0 when P  1 while, for P  1, D has
c                 n
a max.

 A variable coupon rate bond is equivalent to a 1 period zero. Hence its
duration is equal to 1 period (1year, 6 months or the shorter period until
the first coupon’s maturation date. If the bond pays a constant spread s
in addition to the variable coupon rate, its duration is:
n
1  c1 v1  s tvt 
D s                      2
n
1  c1 v1  s vt 
2
It appears that D(s)=1 if s=0. Moreover: dD(s)/ds>0.

Convexity

 The first-order derivative of eq. [3] gives a good approximation for very
small changes in R. The relation between P and R in fact is not linear but
convex:

log P

R
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 The tangent to the curve, whose slope is measured by the modified
D      log P
duration:                 , lies below the curve. Therefore, when R
1 R      R
changes, the modified duration underestimates                         the    rises   and
overestimates the falls in the bond’s price.

 A better approximation can be obtained by the second-order Taylor’s
series expansion:
P      12P
P     R          R 2
R      2 R 2

Hence, by defining the convexity of a bond as:

n      ii  1                n
 xi                            xi ii  11  R
i  2
 P 1 i 1 1  R
2                      i 2
i 1
C                                     
R P
2
P                                        P
we have:
P     D       1
      R  C R
2
P    1 R     2

 Consider 2 bonds having the same duration. When R changes, the price
of the bond with higher convexity increases more or decreases less than
the price of the bond with lower convexity. Hence, it is better to have in
portfolio the more convex bond. But the conclusion is correct for
parallel shifts of the term structure, not in general.

 It is possible to show that: C 
1
1  R       2   D  D   2

  2 where  2 is

a measure of the dispersion of the maturing dates of the cash-flows
around their average given by D.

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Immunization

 When R changes, the effects on capital and income risks have opposite
signs. If the absolute values of the 2 effects are equal, the interest rate
risk is nullified: it has got a complete immunization.

 Assume a flat term structure and consider a bond paying x1 , x2 ,, xn . Its
present value is given by [1]. If all the bond’s cash flows are invested at
R, after m periods (m<n) the accumulation (future value) of the
investment will be:

M m  R  1  R P R
m
[4]

Example

M 2  R  x1 1  R  x 2  x 3 1  R
 3 2
 x n 1  R
n2

2

 1  R x1 1  R
1
 x 2 1  R
2
 x n 1  R
n

 1  R P R
2

 The investment in the bond is made at t=0. Suppose that at t=t1, with
0  t1  m , R has a change but remains unchanged thereafter. Then, if R
goes up, 1  R
m
i.e. the income (reinvestment) effect increases while
P R i.e. the capital effect decreases. If they exactly offset one another,
the future value   M m  R   will be independent of the interest rate changes:

M m  R                            m P R 
 m1  R P R  1  R
m1
[5]                                                  0
R                                    R

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Hence, taking account of [3]: P R / R  1  R          D R P R , it is
1

easily seen that:

M m  R
0             if   m = D(R)
R
which is the conclusion of the Fisher and Weil “classical” immunization:
an m period investment is immunized if it is made in a bond with a
duration equal to m.

 If R changes again after t=t1, it is necessary to adjust the (portfolio)
duration so as to keep it always equal to the time remaining until t=n.

 The assumption of a flat term structure is not necessary. The condition
m=D holds for any term structure provided that it can shift only in a
parallel way (only additive shifts assumption).See Appendix 2.

Immunization of a portfolio of A and L’s: non parallel shifts

 Classical immunization(Fisher-Weil and Redington):

A      1          1
      D A R  C A  R 
2

A    1 R         2
L     1           1
      D L R  C L  R
2

L    1 R         2
  A  L and DA  DL are not sufficient for A  L. It is also
necessary C A  CL .
 With parallel shifts: C A  C L  A  L independently of
the sign of R .

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 Example. Initial situation: flat term structure with          R  5%.
ASSETS                                     LIABILITIES
* 1 year Zero coupon                           * 5 years Zero coupon
(41.1351 Mln, Price= 95.2381)   Mln 39.1763 (100 Mln, Price = 78.3526)   Mln 78.3526
* 9 years Zero coupon
(60.7753 Mln, Price = 64.4609   Mln 39.1763
Total   A                       Mln 78.3526 Total      L                 Mln 78.3526

 At t    0, the term structure curve has an upward parallel shift:
R  55% .
.
ASSETS                                     LIABILITIES
* 1 year Zero coupon                           * 5 years Zero coupon
(41.1351 Mln, Price= 94.7867)   Mln 38.9906 (100 Mln, Price = 76.5134)   Mln 76.5134
*9 years Zero coupon
(60.7753 Mln, Price = 61.7629   Mln 37.5366
Total   A  A                  Mln 76.5272 Total      L  L            Mln 76.5134

AL = 13,800 euro
 Note that  CA  CL produces a gain both in the case of R  0 and
in that of R  0 when the shifts are parallel. If only parallel shifts
were possible, there would be a violation of the non-arbitrage principle.
In fact one could construct a portfolio with 2 zeros(1year and 9 years
and finance it with a zero of 5 years. The duration of assets and
liabilities is the same but C A  CL since the assets’ cash flows are
more dispersed than the liabilities’ ones.

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 If the shift increases the slope of the term structure curve:
R1  5.25%,R5  550%,R9  5.75% , the results are:
.

ASSETS                                      LIABILITIES
* 1 year Zero coupon                           * 5 years Zero coupon
(41.1351 Mln, Price= 95.0119)   Mln 39.0832 (100 Mln, Price = 76.5134)       Mln 76.5134
* 9 years Zero coupon
(60.7753 Mln, Price = 60.4612   Mln 36.7455
Total   A  A                  Mln 75.8287 Total      L  L                Mln 76.5134

AL =  684,700 euro

 For            a        slope          increasing             downward           shift:
R1  4.25%,R5  4.50%,R9  4.75% ,                                  the results are
similar: AL =  761,100 euro.

 The cause of the big losses of the example, when the shifts are slope
increasing, is the large value of the convexity gap CG.

 Minimum risk immunization: To minimize the risk of losses, choose a
portfolio with DA = DL and:         CG  MinCG CG  0 .


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Appendix 1:continuous capitalization
v( t )  1  Rt 
t
Present value of 1 euro to be received at t:
1
m( t )          1  Rt 
t
Future value of 1 euro at t:
v(t )
t
  ( u ) du
Let:    ( t )  ln m( t )                then:         m( t )  e   0

t
   ( u ) du
and:          v(t )  e       0
.
 The present value (price) of a bond paying:                              x1 , x2 ,xn   at:
n
t  1,2, n is given by:W ( t )                          xi v (i ) and its duration is:
i 1
n
 ixi v (i )
D( x )       i 1
n
where        v(t )        represents the market term
 xi v (i )
i 1
structure of interest rates (not necessarily flat).

Appendix 2: Fisher and Weil immunization theorem (any term
 Let  ( t ) be the continuous time TSIR, L  0 an amount to be paid
at   T  0 and x  0 flows to be cashed at t  1,2, n. Let the
present value of assets and liabilities be the same at time t:
n
Wt ( x)       xi v (i )             Lv(T )  Wt ( L)
it
Assume that at time t  t    1 (infinitesimal) there occurs an


additive shift in  ( t ) . Then the post-shift values:
Wt  ( x )  Wt  ( L )               if and only if:     D( X )  T .
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Wt ( x )
Proof:      Let:      Qt ( x , L)              1
Wt ( L)
and let:          (t  , t )   (t )  Y                          for   t  t

For any t up to t it is:
i
   ( u ) du                  T
 xi e     t
1 n            ( u ) du
Qt ( x , L)             T                    i1 xi e    i

   ( u ) du        L
Le     t

The post-shift value of Q will be a function of Y (and             t  ) like:
T                                   T
1 n                 ( u )  Y  du
1 n                             ( u ) du
Qt ( t , x , L , Y )  i1xi e
                               i
 i1 xi e            eY ( T  i ) i
L                    L
with Q  1 for Y  0. Now, calculate the 1st and 2nd derivatives of Q
with respect to Y, to get:
T
1 n                         ( u ) du
Q  (Y )  i1 ( T  i ) xi e       i
eY ( T  i )
L
T
1 n                           ( u ) du
Q  (Y )  i1 ( T  i ) 2 xi e         i
eY ( T  i )  0
L
Since Q ( t )  1 for Y  0 and it is a convex function
 Q (Y )  0, then for any Y  0 it is that Qt  0 if it has a min
for Y  0 i.e. if and only if: Q  ( Y )  0 . This implies that:

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i
n                          ( u ) du        n
 (T     i ) xi e         t                   (T    i ) xi v (i )
i 1
T                             i 1
0
   ( u ) du                         Lv (T )
Le       t

n                          n
T  xi v (i )                 ixi v (i )
i.e.
i 1
 i 1                                  and taking account of the budget
Lv ( T )      Lv ( T )
constraint:

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